Chemical Bonding & Molecular Structure

H = ${1 \over 2}$ (V + M - c + a)

$ \therefore $ H = ${1 \over 2}$(8 + 4) = 6


From structure, it is clear that it has five bond pairs and one lone pair.

Two pi and half sigma bonds are presents in molecule with bond order 2.5.

Moleculer orbital configuration of $N_2^+$ (13 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

Bond order = ${1 \over 2}\left( {9 - 4} \right)$ = 2.5

Molecular orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

And in NO one unpaired electron is present , so it is paramagnetic.

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

And in NO⁺ no unpaired electron is present , so it is diamagnetic.

$Li_2^ + $  (5 electrons) = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^1}}}$

$Li_2^ - $  (7 electrons) = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}$ $\,\sigma _{2{s^1}}^ * \,$

$ \therefore $   Bond order of $Li_2^ + $ = ${{3 - 2} \over 2}$ = 0.5

Bond order of $Li_2^ - $ = ${{4 - 3} \over 2}$ = 0.5

As both $Li_2^ + $ and $Li_2^ - $ has non-zero bond order, so both are stable.

BF₃  $ \to $  BF₄^$-$

NF3 has trigonal pyramidal geometry. Here central atom nitrogen has sp³ hybridization and one lone pair of electron. Because of lone pair there exist repulsion force between lone pair and bond pair of electrons, that is why bond angles are lower than tetrahedral bond angle.

$\therefore\,\,\,$ Total number of lone pair = 9

Note :

According to molecules orbital theory, when a molecule have bond order = 0 then that molecule does not exist.

(a)$\,\,\,$ Configuration of $He_2^{2 + }$ (2 electrons) is = ${\sigma _{1{s^2}}}$

$\therefore\,\,\,$ Bond order = ${1 \over 2}$ (2 $-$0) = 1

(b)$\,\,\,$ Configuration of $He_2^ + $ (3 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^1}}^ * $

$\therefore\,\,\,$ Bond order = ${1 \over 2}$ (2 $-$1) = 0.5

(c) $\,\,\,$ Configuration of $H_2^ - $ (3 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^1}}^ * $

$\therefore$ Bond order = ${1 \over 2}$ (2 $-$ 1) = 0.5

(d) $\therefore\,\,\,$ Configuration of $He_2^{2 - }$ is = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * $

$\therefore\,\,\,$ Bond order = $ = {1 \over 2}$ (2 $-$ 2) = 0

$\therefore\,\,\,$ $H_2^{2 - }$ is not a viable molecule.

Here only KCl is ionic compound all other compound has covalent bond.

This diagram describes antibonding $\pi $ orbital. Here to orbitals laterally overlap to from pi bond.

Here two p orbitals have out of phase combination.

Electrons in $N_2^ + $ = 7 $ \times $ 2 $-$ 1 = 13

Moleculer orbital configuration of $N_2^ + $ (13)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

$\therefore\,\,\,$ $\,{\sigma _{2{p_z}}}$ no. of electrons = 1.

Structure of hydrogen azide is

Bond order of bond I is less than 2 and Bond order of bond II is greater than 2.

(A) BF₃ : sp² Hybridization : Triangular Planar

NF₃ : sp³ Hybridization : Tetrahedral

$CO_3^{2 - }$ : sp² Hybridization : Triangular Planar

(B) $CO_3^{2 - }$ : sp² Hybridization : Triangular Planar

$NO_3^ - $ : sp² Hybridization : Triangular Planar

SO₃ : sp² Hybridization : Triangular Planar

(C) NH₃ : sp³ Hybridization : Tetrahedral

SO₃ : sp² Hybridization : Triangular Planar

$CO_3^{2 - }$ : sp² Hybridization : Triangular Planar

(D) NCl₃ : sp³ Hybridization : Tetrahedral

BCl₃ : sp² Hybridization : Triangular Planar

SO₃ : sp² Hybridization : Triangular Planar

Those species which have unpaired electrons are called paramagnetic species.

(a)   NO⁺ has 14 electrons.

Moleculer orbital configuration of NO⁺ is

${\sigma _{1{s^2}}}$ $\sigma _{1{s^2}}^ * $ ${\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,\, = \,\,{\pi _{2p_y^2}}$

Here is no unpaired electron, So it is Diamagnetic.

(b)    CO has 14 electrons.

Moleculer orbital configuration of CO is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

Here is no unpaired electron so it is diamagnetic.

(c)   $O_2^{2 - }$ has 18 electrons.

Moleculer orbital configuration of $O_2^{2 - }$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * = \,\,\pi _{2p_y^2}^ * $

Here also no unpaired electron present, so it is diamagnetic.

(d)   B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$

Here two unpaired electrons present.

So it is paramagnetic.

Hybridization (X) = ${1 \over 2}$ [ VE + MA – c + a ]

where, VE = No. of valence electrons of central atom
MA = No. of monovalent atoms/groups surrounding the central atom,
c = Charge on the cation,
a = Charge on the anion

(A) [BrF₅] : X = ${1 \over 2}$[ 7 + 5 - 0 + 0] = 6 = sp³d² hybridized

(B) [SF₆] : X = ${1 \over 2}$[ 6 + 6 - 0 + 0] = 6 = sp³d² hybridized

(C) [CrF₆]^3$-$ : Cr⁺³ = [Ar]3d³

(D) [PF₅] : X = ${1 \over 2}$[ 5 + 5 - 0 + 0] = 5 = sp³d hybridized

Note : [CrF₆]^3$-$ shows d²sp³ hybridization not sp³d². So option (C) should also be the correct answer.

Those species which have unpaired electrons are called paramagnetic species.

(a)    CO has 14 electrons.

Moleculer orbital configuration of CO is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

Here is no unpaired electron so it is diamagnetic.

(b)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $

Here 2 unpaired electron present, so it is paramagnetic.

(c)   B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$

Here two unpaired electrons present. So it is paramagnetic.

(d)   NO has 15 electrons.

Moleculer orbital configuration of NO is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $

Here is 1 unpaired electron, So it is Paramagnetic.

CH₄ - sp³ hybridisation - Bond angle (109^o28')

NH₃ - sp³ hybridisation - Bond angle ( 107.8^o)

H₂O - sp³ hybridisation - Bond angle ( 104.5^o)

PH₃ - No hybridisation - Bond angle ( 90^o)

Note : According to DRAGO RULE, lone pair of electrons are in pure s- orbital of the atom P. That is why PH₃ has no hybridisation.

H = ${1 \over 2}$[VE + MA – c + a]

ClF₃ , H = ${1 \over 2}$ (7 + 3 – 0 + 0) = 5 (sp³d)

XeOF₂ , H = ${1 \over 2}$(8 + 2 – 0 + 0) = 5 (sp³d)

XeF₃⁺ , H = ${1 \over 2}$ (8 + 3 – 1 + 0 ) = 5 (sp³d)

All molecules have sp³d hybridization and 3 bond pair + 2 lone pairs. Hence all have identical stucture (T-shape).

Hydrogen bond is a type of strong electrostatic dipole- dipole intersection and dependent on the inverse cube of distance between the molecular ion-dipole -

interaction $ \propto {1 \over {{r^3}}}.$

$H_2^{2+}$ , $He_2$

Both have zero bond order. Thus, they do not exist.

Li₂ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,$

$ \therefore $ Bond order = ${1 \over 2}\left( {4 - 2} \right)$ = 1

$Li_2^+$ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^1}}} \,$

$ \therefore $ Bond order = ${1 \over 2}\left( {3 - 2} \right)$ = 0.5

$Li_2^-$ = ${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}} \,\,\sigma _{1{s^1}}^ * \,$

$ \therefore $ Bond order = ${1 \over 2}\left( {4 - 3} \right)$ = 0.5

The bond order of $Li_2^+$ and $Li_2^-$ is same but $Li_2^+$ is more stable than $Li_2^-$ because $Li_2^+$ is smaller in size and has 2 electrons in antibonding orbitals whereas $Li_2^-$ has 3 electrons in antibonding orbitals. Hence $Li_2^+$ is more stable than $Li_2^-$.

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(A)   $N_2$ has 14 electrons.

Moleculer orbital configuration of $N_2$

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

Here no unpaired electron present, so it is diamagnetic.

(B)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $

Here 2 unpaired electrons present, so it is paramagnetic.

(C)    $S_2$ has 32 electrons.

It also has 2 unpaired electrons like $O_2$, so it is paramagnetic.

(D)   $C_2$ has 12 electrons.

Moleculer orbital configuration of $C_2$

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$

Here no unpaired electron present, so it is diamagnetic.

PF₅ $ \Rightarrow $ sp³d hybridized and shape is trigonal bipyramidal.

BrF₅ $ \Rightarrow $ 5 bond pair and 1 lone pair present so sp³d² hybridized and shape is square pyramidal.

Charge of cation/Size of cation is called polarising power.

$ \therefore $ (i) Polarising power $ \propto $ charge of cation

(ii) Polarising power $ \propto $

1

size of cation

Here the AlCl₃ is satisfying the above two conditions i.e., Al is in +3 oxidation state and also has small size. So it has more covalent character.

The structure of IF₇ pentagonal bipyramidal having sp³d³ hybridisation.

Note :

(1) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

(2) $\,$ Bond order $ = {1 \over 2}$ [N_b $-$ N_a]

N_b = No of electrons in bonding molecular orbital

N_a $=$ No of electrons in anti bonding molecular orbital

(4) $\,\,\,\,$ upto 14 electrons, molecular orbital configuration is



Here N_a = Anti bonding electron $=$ 4 and N_b = 10

(5) $\,\,\,\,$ After 14 electrons to 20 electrons molecular orbital configuration is - - -



Here N_a = 10

and N_b = 10

In O atom 8 electrons present, so in O₂, 8 $ \times $ 2 = 16 electrons present.

Then in $O_2^ + $ no of electrons = 15

in $O_2^ - $ no of electrons = 17

in $O_2^{2 - }$ no of electrons = 18

$\therefore\,\,\,\,$ Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Molecular orbital configuration of O$_2^ {2+ }$ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Molecular orbital configuration of $O_2^ - $ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}$ [ 10 $-$ 8] = 1

As Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

So $O_2^{2+}$ has the shortest bond length.

Number of electron in NO⁺ = number of electron in CN^– = 14 electrons.

As both have same number of electrons so their bond order is equal.

Moleculer orbital configuration of NO⁺ (14 electrons) is

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $

$ \therefore $ B.O = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Moleculer orbital configuration of CN^- (14 electrons) is

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$ \therefore $ B.O = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

In BF₃, B is sp² hybridised and by Back Bonding method strong p$\pi $-p$\pi $ bond is created between filled p-orbital of F and vacant p-orbital of B which leads to shortening of B–F bond length which results in higher bond dissociation energy of the B–F bond. However in CF₄, C does not have any vacant p-orbitals to undergo $\pi $-bonding.

Among F, O and N, F is most electronegative so F pulls bond pair of electron in F - H towards itself and develops highly positive charge on H atom.

This highly positive charged H atom creates stongest hydrogen bonding by taking lone pair of electron form electronegative atom F/N/O. Hence among the given options F – H ...... F is the strongest bond.

Note : F – H ...... N has strongest hydogen bonding among F – H ...... F, F – H ...... O and F – H ...... N because N is least electronegative among F, O and N and can easily donate lone pair of electron.

As charge/size ratio of a cation determines its polarizing power so high charge and small size of the cations increases polarisation.

As the size of the given cations decreases as

K⁺ > Ca²⁺ > Mg²⁺ > Be²⁺

Hence, polarising power decreases as

K⁺ < Ca²⁺ < Mg²⁺ < Be²⁺

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

(a)    $O_2^{2−}$ has 18 electrons.

Moleculer orbital configuration of $O_2^{2−}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

Here is no unpaired electron so it is diamagnetic.

(b)   NO has 15 electrons.

Moleculer orbital configuration of NO is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $

Here is 1 unpaired electron, So it is Paramagnetic.

(c)   $O_2$ has 16 electrons.

Moleculer orbital configuration of $O_2$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^1}^ * $

Here 2 unpaired electron present, so it is paramagnetic.

(d)   $O_2^{+}$ has 15 electrons.

Moleculer orbital configuration of $O_2^{+}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $

Here 1 unpaired electron present, so it is paramagnetic.

(A) Moleculer orbital configuration of $C_2$ (12 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 8

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {8 - 4} \right] = 2$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $C_2^+$ (11 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {7 - 4} \right] = 1.5$

Here 1 unpaired electron present, so it is paramagnetic.

(B) Moleculer orbital configuration of $N_2$ (14 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right] = 3$

Here no unpaired electron present, so it is diamagnetic.

Moleculer orbital configuration of $N_2^+$ (13 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

$\therefore\,\,\,\,$N_a = 4

N_b = 9

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {9 - 4} \right] = 2.5$

Here 1 unpaired electron present, so it is paramagnetic.

(C) Moleculer orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * = \,\,\pi _{2p_y^0}^ * $

$\therefore\,\,\,\,$N_a = 5

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right] = 2.5$

Here is 1 unpaired electron, So it is paramagnetic.

Moleculer orbital configuration of NO⁺ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}}\,= \,{\pi _{2p_y^2}}\,\pi _{2p_x^0}^ * = \,\,\pi _{2p_y^0}^ * $

$\therefore\,\,\,\,$N_a = 4

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right] = 3$

Here is no unpaired electron, So it is diamagnetic.

(D) Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

$\therefore\,\,\,\,$N_a = 6

N_b = 10

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 6} \right] = 2$

Here 2 unpaired electrons present, so it is paramagnetic.

Molecular orbital configuration of O$_2^ + $ (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Here 1 unpaired electrons present, so it is also paramagnetic.

(a)   XeF₄  is sp³d² hybridised with 4 bond pairs and 1 lone pair and structure is square planar. Here all the bond lengths are equal.



(b) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral. Here all the bond lengths are equal.



(c)   SF₄ is sp³d hybridised with 4 bond pairs and 1 lone pair and its expected trigonal bipyramidal geometry gets distorted due to presence of a lone pair of electrons and it becomes distorted tetrahedral or see-saw with the bond angles equal to < 120^o and 179^o instead of the expected angles of 120^o and 180^o respectively. Here axial and equitorial both bonds are presents. And we know axial bonds are longer and weaker.



(d) SiF₄ is sp³ hybridisation and regular tetrahedral geometry. Here all the bond lengths are equal.

(A) Molecular orbital configuration of O $_2^{2 - }$ (18 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ * $

So O $_2^{2 - }$ has no unpaired electrons.

(B) Molecular orbital configuration of B₂ (10 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^1}} = {\pi _{2p_y^1}}$

Here in B₂, 2 unpaired electrons present.

(C) Moleculer orbital configuration of $N_2^{ + }$ (13 electrons)

= ${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^1}}$

Here in $N_2^{ + }$, 1 unpaired electron present.

(D) Molecular orbital configuration of O₂ (16 electrons) is

${\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,$ ${\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,$ ${\sigma _{2p_z^2}}\,\,{\pi _{2p_x^2}} = {\pi _{2p_y^2}}\,\,\pi _{2p_x^1}^ * \,\, = \pi _{2p_y^1}^ * $

So O$_2$ has 2 unpaired electrons.

Nitrogen(N) atom is smaller in size so it's lone pair is relatively unstable. That is why repulsion between lone pairs present on nitrogen atom and bonded pairs of electrons is relatively high.

But size of Sb is more so its lone pair is relatively stable. That is why it does not want to perticipate in bond pair lone pair repulsion and decreases lp-bp repulsion.

The lattice energy of an ionic compound is a measure of the strength of the bonds in that ionic compound. Specifically, it is the energy required to separate one mole of an ionic solid into its constituent ions in the gaseous state. The lattice energy depends on several factors, primarily the charge on the ions and the size (or radius) of the ions.

1. Charge on the ions: The lattice energy is directly proportional to the product of the charges of the cation and the anion. Higher charges result in stronger electrostatic attraction between the ions, thereby increasing the lattice energy. This relationship can be derived from Coulomb's law, which states that the force of attraction between two charged particles is directly proportional to the product of their charges.

Mathematically:

$U \propto \frac{q_1 \cdot q_2}{r}$

where $U$ is the lattice energy, $q_1$ and $q_2$ are the charges on the ions, and $r$ is the distance between the centers of the ions.

2. Size of the ions: The lattice energy is inversely proportional to the distance between the ions, which depends on the sum of their ionic radii. Smaller ions can come closer together, increasing the electrostatic attractions between them, thereby increasing the lattice energy. Larger ions, being further apart, result in a lower lattice energy.

Considering these two factors, the correct choice is:

Option D: Charge on the ion and size of the ion

TIPS/Formulae :

A diamagnetic substance contains no unpaired electron.

${H_2}$ is diamagnetic as it contains all paired electrons

$\mathop {{H_2} = \sigma _b^2}\limits_{\left( {diamagnetic} \right)} \,\,,\,\,\mathop {H_2^ + = \sigma _b^1,}\limits_{\left( {paramagnetic} \right)} \,\,\mathop {H_2^ - = \sigma _b^2,}\limits_{\left( {paramagnetic} \right)} $

$\mathop {\sigma _a^{ * 1};He_2^ + }\limits_{\left( {paramagnetic} \right)} = \mathop {\sigma _b^2,\sigma _a^{ \ne 1}}\limits_{\left( {paramagnetic} \right)} $

Here eight 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1, $\angle $5M1, $\angle $5M2, $\angle $5M3, $\angle $5M4.

Here twelve 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1, $\angle $5M1, $\angle $5M2, $\angle $5M3, $\angle $5M4, $\angle $6M1, $\angle $6M2, $\angle $6M3, $\angle $6M4.

Here four 90^o angles between bond pair and bond pair. Those angles are $\angle $1M2, $\angle $2M3, $\angle $3M4, $\angle $4M1.

Here six 90^o angles between bond pair and bond pair. Those angles are $\angle $1M3, $\angle $1M4, $\angle $1M5, $\angle $2M3, $\angle $2M4, $\angle $2M5.

Structure of  H₃BO₃  is



Here Boron has 3 sigma bond. So it is sp² hybridised.

And oxygen has two sigma bond and two lone pair. So it sp³ hybridised.

1. In H₂S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H₂S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(S) atom and size of S^$-$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

2. In NH₃, 3 Bond pair(BP) + 1 lone pair (LP) present so angle between bond 107^o.



3. $\,\,\,\,$ BF₃ has sp² hybridization. So bond angle is 120^o.



4.
Here Si is sp³ hybridised and shape is regular tetrahedral and bond angle 109^o 28'

Molecular orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ * $

$\therefore\,\,\,\,$ N_b = 10

N_a = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Similarly Molecular orbital configuration of NO⁺ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,$

$\therefore\,\,\,\,$ N_b = 10

N_a = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

Note :
(2) $\,\,\,\,$ Bond length $ \propto $ ${1 \over {Bond\,\,order}}$

So, bond length in NO > NO⁺

Regular Tetrahedral structure is possible in sp³ hybridization where central atom has 4 bond pair and no lone pair.

(a)   XeF₄  is sp³d² hybridised and structure is square planar.



(b)   [Ni(CN)₄]^$-$2 is coordinate compound and oxidation number of Ni is +2.

Electronic configuration of Ni⁺² is $=$ [Ar]3d⁸



But because of CN^$-$ ion which is a strong field ligand , it can perform pairing of electron.



And the structure of dsp² hybridization is square planar.



(C) $\,\,\,\,$ BF$_4^ - $, 4 bond pair present so angle is 109^o 28^' and sp³ hybridised. So structure is regular tetrahedral.



(d)   

SF₄ is sp³d hybridised and structure is see-saw.

Here $\mu $_total = $\mu $₁ + $\mu $₂ $ \ne $ 0

$\mu $_total = $\mu $₁ $-$ $\mu $₂ = 0

$\mu $_total = $\mu $₁ + $\mu $₂ $ \ne $ 0

$\mu $₁ + $\mu $₂ + $\mu $₃ = $\mu $₄

   $\mu $_total = 0

In NO₂ and O₃ have permanent dipole moment as both of them are not symmetric molecule.

In CO₂ and SiF₄ each individual bond c = O and S- $-$ F have dipole moment but because of symmetric structure individual dipole moment get's cancelled.

(a)   H₂O is sp³ hybridized, and oxygen atom has 2 bond pair and 2 lone pair. So the angle between two O $-$ H bond is 104.5^o

(b)   In H₂S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H₂S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(s) atom and size of S^$-$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

(c)   NH₃ has sp³ hybridization and N atom has 3 bond pair and one lone pair so bond angle is 107^o

(d)   SO₂ is sp² hybridized and bond angle is 117^o.